Problem 56 - Essential Calculus Early Transcendentals

Question

If \(f^{\prime \prime}\) is continuous, show that $$\quad \lim _{h \rightarrow 0} \frac{f(x+h)-2 f(x)+f(x-h)}{h^{2}}=f^{\prime \prime}(x)$$

Step-by-Step Solution

Answer

The limit is \( f''(x) \).

Step 1: Define the Expression

We start with the expression \( \frac{f(x+h)-2f(x)+f(x-h)}{h^2} \) and aim to prove that its limit as \( h \) approaches zero equals \( f''(x) \).

Step 2: Use Taylor Expansion

Apply the Taylor series expansion for \( f(x+h) \) and \( f(x-h) \). This gives:\[f(x+h) = f(x) + f'(x)h + \frac{f''(x)h^2}{2} + R_1(h)\]\[f(x-h) = f(x) - f'(x)h + \frac{f''(x)h^2}{2} + R_2(h)\]where \( R_1(h) \) and \( R_2(h) \) are the remainder terms.

Step 3: Substitute the Taylor Expansions

Replace \( f(x+h) \) and \( f(x-h) \) in the original expression.\[\frac{(f(x) + f'(x)h + \frac{f''(x)h^2}{2} + R_1(h)) - 2f(x) + (f(x) - f'(x)h + \frac{f''(x)h^2}{2} + R_2(h))}{h^2}\]

Step 4: Simplify the Expression

Simplify the numerator:\[ (f(x) + f'(x)h + \frac{f''(x)h^2}{2} + R_1(h)) + (f(x) - f'(x)h + \frac{f''(x)h^2}{2} + R_2(h)) - 2f(x) = f''(x)h^2 + R_1(h) + R_2(h)\]Then divide by \( h^2 \):\[\frac{f''(x)h^2 + R_1(h) + R_2(h)}{h^2} = f''(x) + \frac{R_1(h)+R_2(h)}{h^2}\]

Step 5: Take the Limit as h Approaches 0

Since \( f''(x) \) is continuous and the remainder terms \( R_1(h) \), \( R_2(h) \) go to zero faster than \( h^2 \), their contributions vanish:\[\lim_{h \to 0} \left( \frac{R_1(h) + R_2(h)}{h^2} \right) = 0\]Thus, \[\lim_{h \to 0} \left( f''(x) + \frac{R_1(h)+R_2(h)}{h^2} \right) = f''(x)\]

Step 6: Conclusion

The limit of the expression as \( h \) approaches zero is indeed \( f''(x) \), proving the result.

Key Concepts

Taylor Series Expansion

The Taylor series expansion is a powerful mathematical tool used to approximate complex functions using polynomials. This is especially helpful in understanding the behavior of functions near a specific point. When you expand a function into a Taylor series, you express it as a sum of its derivatives taken at a point, multiplied by powers of the distance from that point.

• For a function \( f(x) \), its Taylor series expansion around a point \( a \) is given by:
- \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_n(x) \]

This series continues infinitely, but for practical applications, it is typically truncated after a few terms.
In the context of the given problem, we use the Taylor expansion at the point \( x \) to express \( f(x+h) \) and \( f(x-h) \). The first few terms of the series for these expansions show the values of the function at and near \( x \), helping to approximate changes in function values with increasing or decreasing \( h \).
Understanding the remainder terms \( R_1(h) \) and \( R_2(h) \) is crucial. These terms account for the error between the exact function value and its polynomial approximation. As \( h \) approaches zero, these remainder terms become negligible, ensuring the accuracy of the approximation.

Second Derivative

The second derivative of a function tells us about the concavity and the inflection points of the function. It provides insights into how the slope of the function is changing, indicating whether the function is curving upwards or downwards.

• Mathematically, the second derivative is the derivative of the first derivative. If \( f'(x) \) was a function describing the slope, then \( f''(x) \) reveals how that slope itself changes.

In the given problem, we are tasked with proving that the limit of a specific expression as \( h \) approaches zero equals the second derivative \( f''(x) \).
The expression \( \frac{f(x+h)-2f(x)+f(x-h)}{h^2} \) is essentially a symmetric difference quotient, and as \( h \) tends to zero, it approaches \( f''(x) \). This is because symmetric differences take into account the slope changes on both sides of a point, thus approximating the second derivative effectively.
Therefore, due to the continuity of \( f'' \), this expression provides a consistent means of evaluating the curvature of the function \( f \) at the given point \( x \) using limits.

Continuity of Functions

Continuity is a fundamental concept that describes how a function behaves and whether it can be plotted without lifting your pen from the paper. A continuous function has no breaks, jumps, or holes. This property is essential for many calculus operations, including taking derivatives.

• A function \( f \) is continuous at a point \( c \) if the limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \).

This ensures that as \( x \) closes in on \( c \), the function's values do not unexpectedly change.
In the context of the problem, the continuity of the second derivative \( f''(x) \) is crucial. It guarantees that the behavior of \( f''(x) \) is predictable and without sudden jumps, making our calculations involving limits more reliable. When derivatives are continuous, you avoid potential pitfalls such as cusps or vertical tangents, ensuring that limit operations will smoothly converge to the true derivative value.
As you work towards proving limits such as \( \lim_{h \to 0} \frac{R_1(h) + R_2(h)}{h^2} = 0 \), the continuity of \( f''(x) \) helps simplify assumptions about the behavior of remainder terms, verifying their insignificant contributions at very small \( h \). This allows the primary term \( f''(x) \) in our Taylor expansion to become the limit value effectively.